Using the fact that $\sin (A + B) = \sin A\cos B + \cos A\sin B$ and the differentiation, obtain the sum formula for cosines.

$\sin (A + B) = \sin A\cos B + \cos A\sin B$ ....(i)

Consider A and B as function of t and differentiating both sides of (i) w.r.t. t, we get

$\cos (A + B)\left( {\cfrac{{dA}}{{dt}} + \cfrac{{dB}}{{dt}}} \right)$

$= \sin A( - \sin B)\cfrac{{dB}}{{dt}} + \cos B\left[ {\cos A\cfrac{{dA}}{{dt}}} \right] + \cos A\cos B\cfrac{{dB}}{{dt}} + \sin B( - \sin A)\cfrac{{dA}}{{dt}}$

$\Rightarrow$ $\cos (A + B)\left( {\cfrac{{dA}}{{dt}} + \cfrac{{dB}}{{dt}}} \right)$

$= (\cos A\cos B - \sin A\sin B)\left( {\cfrac{{dA}}{{dt}} + \cfrac{{dB}}{{dt}}} \right)$

$\Rightarrow$ $\cos (A + B) = \cos A\cos B - \sin A\sin B.$